How do you find the projection of u onto v given #u=<2,2># and #<v=6,1>#?

1 Answer
Oct 25, 2016

Answer:

The vector projection of #vecu# onto #vecv# #=〈84/37,14/37〉#

Explanation:

The vector projection of #vecu# over #vecv# is given by
#=(vecu.vecv)/(∣vecv∣^2)vecv#

Dot product #vecu.vecv=〈u_1,u_2〉.〈v_1,v_2〉=u_1v_1+u_2v_2#
The dot product #vecu.vecv=〈2,2〉.〈6,1〉=12+2=14#

#∣vecv∣^2=〈v_1,v_2〉〈v_1,v_2〉=v_1^2+v_2^2#
So #∣vecv∣^2=〈6,1〉〈6,1〉=36+1=37#

And finally the projection is #=14/37〈6,1〉=〈84/37,14/37〉#